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Chapter 9: Problem 61

Verify that $$ \frac{\partial^{2} f}{\partial x \partial y}=\frac{\partial^{2} f}{\partial y \partial x} $$ in the cases (a) \(f(x, y)=x^{2} \cos y\) (b) \(f(x, y)=\sinh x \cos y\)

Short Answer

Expert verified
The mixed partial derivatives are equal for both cases (a) and (b).

Step by step solution

01

Calculate First Partial Derivative with Respect to x

For part (a), the given function is \( f(x, y) = x^2 \cos y \). First, we find the partial derivative with respect to \( x \): \[\frac{\partial f}{\partial x} = 2x \cos y.\] For part (b), where \( f(x, y) = \sinh x \cos y \), the first partial derivative with respect to \( x \) is: \[\frac{\partial f}{\partial x} = \cosh x \cos y.\]
02

Calculate Second Partial Derivative with Respect to y

Continuing with part (a), we now find the partial derivative of \( \frac{\partial f}{\partial x} = 2x \cos y \) with respect to \( y \):\[\frac{\partial^2 f}{\partial x \partial y} = -2x \sin y.\]For part (b), take the derivative of \( \frac{\partial f}{\partial x} = \cosh x \cos y \) with respect to \( y \):\[\frac{\partial^2 f}{\partial x \partial y} = -\cosh x \sin y.\]
03

Calculate First Partial Derivative with Respect to y

For part (a), differentiate \( f(x, y) = x^2 \cos y \) with respect to \( y \): \[\frac{\partial f}{\partial y} = -x^2 \sin y.\]For part (b), differentiate \( f(x, y) = \sinh x \cos y \) with respect to \( y \):\[\frac{\partial f}{\partial y} = -\sinh x \sin y.\]
04

Calculate Second Partial Derivative with Respect to x

Now, for part (a), take the derivative of \( \frac{\partial f}{\partial y} = -x^2 \sin y \) with respect to \( x \):\[\frac{\partial^2 f}{\partial y \partial x} = -2x \sin y.\]For part (b), differentiate \( \frac{\partial f}{\partial y} = -\sinh x \sin y \) with respect to \( x \): \[\frac{\partial^2 f}{\partial y \partial x} = -\cosh x \sin y.\]
05

Verify Equality of Mixed Second Derivatives

For both part (a) and part (b), the mixed second partial derivatives are equal. In part (a): \[\frac{\partial^2 f}{\partial x \partial y} = -2x \sin y = \frac{\partial^2 f}{\partial y \partial x}.\]In part (b):\[\frac{\partial^2 f}{\partial x \partial y} = -\cosh x \sin y = \frac{\partial^2 f}{\partial y \partial x}.\] This confirms the equality of the mixed second derivatives \( \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} \) for both functions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mixed Second Derivatives
Mixed second derivatives are important in multivariable calculus. They involve taking the second partial derivative of a function, first with respect to one variable, and then with respect to another. For a function \( f(x, y) \), the mixed second derivative \( \frac{\partial^2 f}{\partial x \partial y} \) means you first differentiate with respect to \( x \), then with respect to \( y \). Similarly, \( \frac{\partial^2 f}{\partial y \partial x} \) means you first differentiate with respect to \( y \), then \( x \). Mixed derivatives can provide information about how a surface defined by \( f(x, y) \) changes in different directions.
Partial Derivative with Respect to x
A partial derivative with respect to \( x \) measures how a function changes as \( x \) changes, while keeping \( y \) constant. It's like slicing a 3D surface along a plane parallel to the \( xz \)-plane.For a function \( f(x, y) \), the partial derivative \( \frac{\partial f}{\partial x} \) involves differentiating \( f \) with respect to \( x \), treating \( y \) as a constant.
  • In exercise (a) for \( f(x, y) = x^2 \cos y \), we find \( \frac{\partial f}{\partial x} = 2x \cos y \).
  • For exercise (b) with \( f(x, y) = \sinh x \cos y \), \( \frac{\partial f}{\partial x} = \cosh x \cos y \).
Partial Derivative with Respect to y
The partial derivative with respect to \( y \) shows how a function varies as \( y \) changes, keeping \( x \) fixed. This is like slicing the surface along a plane parallel to the \( yz \)-plane.To find \( \frac{\partial f}{\partial y} \) for function \( f(x, y) \), we differentiate \( f \) considering \( x \) constant.
  • In part (a) for \( f(x, y) = x^2 \cos y \), the derivative is \( \frac{\partial f}{\partial y} = -x^2 \sin y \).
  • In part (b), \( f(x, y) = \sinh x \cos y \), hence \( \frac{\partial f}{\partial y} = -\sinh x \sin y \).
Equality of Mixed Derivatives
The equality of mixed derivatives, also known as Clairaut's theorem, is a fundamental property in calculus. It states that for most well-behaved functions, the order of differentiation does not matter; thus \( \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} \).This property holds if the function \( f(x, y) \) is continuous and has continuous second partial derivatives. In our exercises, we verified this theorem:
  • For (a), \( f(x, y) = x^2 \cos y \), both mixed derivatives gave \( -2x \sin y \).
  • For (b), \( f(x, y) = \sinh x \cos y \), both yielded \( -\cosh x \sin y \).
This unanimous result confirms the swaps of derivative orders leads to identical results.

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Most popular questions from this chapter

By means of sketches of the graphs \(y=1 / x\) and \(y=\tan x\), show that the equation \(x \tan x=1\) has a root between \(x=0\) and \(x=\frac{1}{2} \pi\) and an infinity of roots near \(x=k \pi\), where \(k=1,2,3, \ldots .\) Deduce which of the two iterations (a) \(x_{n+1}=\cot x_{n}\) (b) \(x_{n+1}=\tan ^{-1}\left(1 / x_{n}\right)+k \pi\) is convergent to the roots, and use it to locate the smallest positive root to \(6 \mathrm{dp}\).

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Verify that\(z=\ln \sqrt{\left(x^{2}-y^{2}\right)}\) satisfies the equation $$ \left(\frac{\partial z}{\partial x}\right)^{2}+\frac{\partial^{2} z}{\partial y \partial x}+\left(\frac{\partial z}{\partial y}\right)^{2}=\frac{1}{(x-y)^{2}} $$

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